How do you find the zeros, real and imaginary, of $y = - x^{2} - 12 x - 8$ $y= -x^2-12x-8$ using the quadratic formula?

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How do you find the zeros, real and imaginary, of $y = - x^{2} - 12 x - 8$ $y= -x^2-12x-8$ using the quadratic formula?

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Answer 1 · 2017-01-31T23:23:31

$x = - 6 + 2 \sqrt{7};$ $x=-6+2sqrt7;$ or $x = - 6 - 2 \sqrt{7}$ $x=-6-2sqrt7$

Explanation:

When using the quadratic formula, the zeros are the two values for $x$ $x$ . When solving quadratic equations, the equation must be equal to $0$ $0$ .

Your equation: $y = - x^{2} - 12 x - 8$ $y=-x^2-12x-8$

Set $y$ $y$ equal to zero.

$0 = - x^{2} - 12 x - 8$ $0=-x^2-12x-8$

Multiply by $- 1$ $-1$ to make $a$ $a$ positive.

$x^{2} + 12 x \pm 8$ $x^2+12x+-8$

$- x^{2} - 12 x - 8$ $-x^2-12x-8$ is a quadratic equation in the form of $a^{2} + b x + c$ $"color(red)(a)^2+color(blue)bx+color(green)(c)$ , where the coefficients are $a = (1)$ $"color(red)(a)=(1)$ , $b = 12$ $color(blue)(b)=12$ , $c = 8$ $color(green)(c)=8$ .

Quadratic equation

$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $x=(color(blue)(-b)+-sqrt(color(blue)(b)^2-4color(red)(a)color(green)(c)))/(2color(red)a)$

Substitute the known values into the equation.

$x = \frac{- (12) \pm \sqrt{12^{2} - 4 (1) (8)}}{(2 \cdot 1)}$ $x=("-(color(blue)(12))+-sqrt(color(blue)(12)^2-4(color(red)(1))(color(green)(8))))/((2*color(red)(1)))$

$x = \frac{- 12 \pm \sqrt{144 - (32)}}{- 2}$ $x=(-12+-sqrt(144-(32)))/(-2)$

$= \frac{- 12 \pm \sqrt{112}}{- 2}$ $=(-12+-sqrt(112))/(-2)$

Factor $\sqrt{112}$ $sqrt112$ .

$\sqrt{112} = \sqrt{16 \times 7} = 4 \sqrt{7}$ $sqrt112=sqrt(16xx7)=4sqrt7$

$x = \frac{- 12 \pm 4 \sqrt{7}}{2}$ $x=(-12+-4sqrt7)/2$

Simplifyby factoring out $2$ $2$ in $- 12, \pm 4, and 2$ $-12, +-4, and 2$ .

$x = - 6 \pm 2 \sqrt{7}$ $x=-6+-2sqrt7$

$x = - 6 + 2 \sqrt{7};$ $x=-6+2sqrt7;$ or $x = - 6 - 2 \sqrt{7}$ $x=-6-2sqrt7$

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How do you find the zeros, real and imaginary, of $y = - x^{2} - 12 x - 8$ $y= -x^2-12x-8$ using the quadratic formula?

1 Answer

Explanation:

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Science

Math

Humanities

... and beyond

How do you find the zeros, real and imaginary, of y=−x2−12x−8y= -x^2-12x-8 using the quadratic formula?

1 Answer

Explanation:

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Impact of this question

How do you find the zeros, real and imaginary, of $y = - x^{2} - 12 x - 8$ $y= -x^2-12x-8$ using the quadratic formula?